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Functions With Intervals

  1. Nov 10, 2012 #1
    1. The problem statement, all variables and given/known data

    If A and B are sets and f : A → B, then for any subset S of A we define:

    f(S) = {b ∈ B : b = f(a) for some a ∈ S}

    Similarly, for any subset T of B we define the pre-image of T as:

    f[itex]^{-1}[/itex](T) = {a ∈ A : f(a) ∈ T}

    Note that f[itex]^{-1}[/itex](T) is well defined even if f does not have an inverse.

    Now let f : R → R be defined as f(x) = [itex]x^{2}[/itex]

    Let S1 denote the closed interval [−2, 1], and let S2 be the open interval (−1, 2). Also let T1 = S1 and T2 = S2.

    Determine:

    f(S1 ∪ S2)

    f(S1) ∪ f(S2)

    f(S1 ∩ S2)

    f(S1) ∩ f(S2)

    f[itex]^{-1}[/itex](T1 ∪ T2)

    f[itex]^{-1}[/itex](T1) ∪ f[itex]^{-1}[/itex](T2)

    f[itex]^{-1}[/itex](T1 ∩ T2)

    f[itex]^{-1}[/itex](T1) ∩ f[itex]^{-1}[/itex](T2)


    3. The attempt at a solution

    I think i'm being asked to verify that each subset satisfies "some b = f(a) for some a ∈ S" where S is the subset i have to compute.

    For example, the first question is f(S1 U S2) = f( [-2,2) ).

    My question is, what do i have to do next? I can't plug in every value x where -2 =< x < 2 since i'm using real numbers.
     
    Last edited: Nov 10, 2012
  2. jcsd
  3. Nov 10, 2012 #2

    haruspex

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    Yes.
    Of course you can't, but if you sketch a graph of the function it will suggest an answer which you can then set about proving.
     
  4. Nov 10, 2012 #3
    Well since my first answer has a range spanning from -2 up to but not including 2, it is one-to-one. The question wants me to answer whether or not every value passed to the function will map to one value, i think.

    It's easy for me to see that f( [-2,2) ) is valid, but how would i go about showing it? The question doesn't explicitly ask me to solve anything, it just says "determine" so maybe the answer is just f( [-2,2) )... Bah, now i'm all paranoid that i'm missing something...
     
  5. Nov 10, 2012 #4

    haruspex

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    How so? 1-1 would mean that no two values in [-2,2) get mapped by f to the same value. False for f(x)=x2, surely.
    No, it's asking for the set of values obtained by applying the function to all the numbers in the range [-2,2).
     
  6. Nov 11, 2012 #5
    Could i denote the set of values as [0,4]? Since we're working with real numbers it wouldn't be feasible to list all of the possible values.
     
  7. Nov 11, 2012 #6

    micromass

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    Yes. That is the correct answer. But can you justify it?? Can you actually prove that [itex]f([-2,2])=[0,4][/itex]??
     
  8. Nov 11, 2012 #7
    Would it be sufficient to say that for every value x in (-2,2], f(x) will be no less than 0 and no greater than 4? I mean, there doesn't seem to be any other way to put it. I can't invert the function and plug in 0 and 4, so all i can really do is plug in the important values in my domain (ie. -1.99999, 0, 2)

    I finished the rest of them...

    f(S1) U f(S2) = f( [-2,1] ) U f( (-1,2) ) = [1,4] U (1,4) = [1,4]

    f(S1 ∩ S2) = f( (-1,1] ) = [0,1]

    f(S1) ∩ f(S2) = f( [-2,1] ) ∩ f( (-1,2) ) = [1,4] ∩ (1,4) = (1,4)

    I'm starting to wonder whether i have to actually prove my answer for each question, it seems like it would be a bit redundant... Maybe i am just supposed to show the range and that's it?

    The second part with the inverse is confusing to me.

    f'(T1 U T2) = f'( [-2,2) )

    So the result of this would be every number x where [itex] x^{2} \in [-2,2) [/itex] or in other words, the square roots of every number in [-2,2), i think. Which would give me a messy result.
     
    Last edited: Nov 11, 2012
  9. Nov 11, 2012 #8

    haruspex

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    That shows f(S1 U S2) [itex]\subseteq[/itex] [0,4]. Also need to show the converse.
    No. Please do sketch the graph in each case. What does f look like over [-2,1]?
    Yes.
    No.
    Every real number x, that is
    No, that's not the same. There are two differences. First, you don't have to have a square root for every number. There is no real number for which the square is -1, so the answer will not include a square root of -1. Secondly, 'the square root' of a positive number is generally defined to mean the positive square root, but here we want every real number that has a square in the desired range.
    This question illustrates the difference between a pre-image function, as can be applied to sets, and an inverse function as applied to the values.
     
  10. Nov 11, 2012 #9
    So the answer to f[itex]^{-1}[/itex]( [-2,2) ) is the set of real numbers obtained through [itex]x^{2}[/itex] in [-2,2)

    Would i do this with set builder notation? The answer can't be [0,1.4142]..
     
  11. Nov 12, 2012 #10

    haruspex

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    No, but that's nearly right. Check the bounds.
     
  12. Nov 12, 2012 #11
    So if i were to write it as [0, √2], would that technically be correct? This solution doesn't account for negative numbers, which throws me off.
     
  13. Nov 12, 2012 #12

    haruspex

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    As I said, that's nearly right. Are the squares of all numbers in that range contained within [-2,2)?
     
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